C

Explanation

Total Contestants = 6

Condition Contestants= 5
Reason==>No Hope always finishes before Goldenrod, Consider both as one person as their order is N G, means they will b together all the times

So, the possible arrangement for 6 people in which No Hope always finishes before Goldenrod = 5! = 120

OA

two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

800guy wrote:Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
(A) 720
(B) 360
(C) 120
(D) 24
(E) 21

what is the OA? is it C

real2008 wrote:

800guy wrote:Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
(A) 720
(B) 360
(C) 120
(D) 24
(E) 21

what is the OA? is it C

Nope. OA is [spoiler][/spoiler].

This problem is similar to 6 mobsters who go to watch a movie where the condition is one person has to stay behind the other person in a queue to get the movie ticket to keep an eye on the other.

The answer is very well explained.

800guy wrote:Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
(A) 720
(B) 360
(C) 120
(D) 24
(E) 21

Common sense and logic go a long way to solving combinatorics problems on the GMAT.

If we didn't care about the order of G and N, there would be 6! = 720 possible arrangements of horses.

However, we do care about the order of G and N; we know that N always finishes before G does.

So, let's use a tiny bit of common sense. Assuming that there are no ties (something which should have been explicitly stated in the question, by the way - otherwise the question is waaaay more complicated and the correct answer isn't even present), N will finish ahead of G in 50% of the possible arrangements (if it wasn't exactly 50%, that would mean that G finishes ahead of N more or less than half the time, which makes no sense).

Therefore, the correct answer is simply 6! * (1/2) = 720/2 = 360... choose B.

lalitaroral wrote:C

Explanation

Total Contestants = 6

Condition Contestants= 5
Reason==>No Hope always finishes before Goldenrod, Consider both as one person as their order is N G, means they will b together all the times

So, the possible arrangement for 6 people in which No Hope always finishes before Goldenrod = 5! = 120

You've assumed that N must finish immediately before G, which isn't explicit in the question. Based on the information we have, we only know that N finishes some time before G does.

Find question #65 after clicking the link below

OA is B- 360

This breaks down the solution, pretty much what 800guy did

https://www.scribd.com/doc/7117835/GMAT- ... h-Problems

J

uw490 wrote:Find question #65 after clicking the link below

OA is B- 360

This breaks down the solution, pretty much what 800guy did

https://www.scribd.com/doc/7117835/GMAT- ... h-Problems

J

That app keeps crashing IE for me!

In any case, solving for 6!/2 seems just a mite quicker .